f'(x)=2x-3-1/(x-1)?
Supongamos que f'(x)=0:
2x-3-1/(x-1)? =0
(2x-3)(x-1)? =1
2x? -7x? 8x-4=0
(x-2)(2x?-3x 2)=0
Entonces x=2.
∵1 lt; x lt2, f' (x) < 0
x gt2 punto, f '(x)>; f(2)=4-6 1 3=2
f'(x)=2x-3-1/(x-1)?
Supongamos que f'(x)=0:
2x-3-1/(x-1)? =0
(2x-3)(x-1)? =1
2x? -7x? 8x-4=0
(x-2)(2x?-3x 2)=0
Entonces x=2.
∵1 lt; x lt2, f' (x) < 0
x gt2 punto, f '(x)>; f(2)=4-6 1 3=2